# 1649. Create Sorted Array through Instructions

## Description

Given an integer array instructions, you are asked to create a sorted array from the elements in instructions. You start with an empty container nums. For each element from left to right in instructions, insert it into nums. The cost of each insertion is the minimum of the following:

• The number of elements currently in nums that are strictly less than instructions[i].
• The number of elements currently in nums that are strictly greater than instructions[i].

For example, if inserting element 3 into nums = [1,2,3,5], the cost of insertion is min(2, 1) (elements 1 and 2 are less than 3, element 5 is greater than 3) and nums will become [1,2,3,3,5].

Return the total cost to insert all elements from instructions into nums. Since the answer may be large, return it modulo 109 + 7

Example 1:

Input: instructions = [1,5,6,2]
Output: 1
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 5 with cost min(1, 0) = 0, now nums = [1,5].
Insert 6 with cost min(2, 0) = 0, now nums = [1,5,6].
Insert 2 with cost min(1, 2) = 1, now nums = [1,2,5,6].
The total cost is 0 + 0 + 0 + 1 = 1.

Example 2:

Input: instructions = [1,2,3,6,5,4]
Output: 3
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 2 with cost min(1, 0) = 0, now nums = [1,2].
Insert 3 with cost min(2, 0) = 0, now nums = [1,2,3].
Insert 6 with cost min(3, 0) = 0, now nums = [1,2,3,6].
Insert 5 with cost min(3, 1) = 1, now nums = [1,2,3,5,6].
Insert 4 with cost min(3, 2) = 2, now nums = [1,2,3,4,5,6].
The total cost is 0 + 0 + 0 + 0 + 1 + 2 = 3.


Example 3:

Input: instructions = [1,3,3,3,2,4,2,1,2]
Output: 4
Explanation: Begin with nums = [].
Insert 1 with cost min(0, 0) = 0, now nums = [1].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3].
Insert 3 with cost min(1, 0) = 0, now nums = [1,3,3,3].
Insert 2 with cost min(1, 3) = 1, now nums = [1,2,3,3,3].
Insert 4 with cost min(5, 0) = 0, now nums = [1,2,3,3,3,4].
​​​​​​​Insert 2 with cost min(1, 4) = 1, now nums = [1,2,2,3,3,3,4].
​​​​​​​Insert 1 with cost min(0, 6) = 0, now nums = [1,1,2,2,3,3,3,4].
​​​​​​​Insert 2 with cost min(2, 4) = 2, now nums = [1,1,2,2,2,3,3,3,4].
The total cost is 0 + 0 + 0 + 0 + 1 + 0 + 1 + 0 + 2 = 4.


Constraints:

• 1 <= instructions.length <= 105
• 1 <= instructions[i] <= 105

## Solutions

Binary Indexed Tree or Segment Tree.

• class BinaryIndexedTree {
private int n;
private int[] c;

public BinaryIndexedTree(int n) {
this.n = n;
this.c = new int[n + 1];
}

public void update(int x, int v) {
while (x <= n) {
c[x] += v;
x += x & -x;
}
}

public int query(int x) {
int s = 0;
while (x > 0) {
s += c[x];
x -= x & -x;
}
return s;
}
}

class Solution {
public int createSortedArray(int[] instructions) {
int m = 0;
for (int x : instructions) {
m = Math.max(m, x);
}
BinaryIndexedTree tree = new BinaryIndexedTree(m);
int ans = 0;
final int mod = (int) 1e9 + 7;
for (int i = 0; i < instructions.length; ++i) {
int x = instructions[i];
int cost = Math.min(tree.query(x - 1), i - tree.query(x));
ans = (ans + cost) % mod;
tree.update(x, 1);
}
return ans;
}
}

• class BinaryIndexedTree {
public:
BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) {}

void update(int x, int delta) {
while (x <= n) {
c[x] += delta;
x += x & -x;
}
}

int query(int x) {
int s = 0;
while (x) {
s += c[x];
x -= x & -x;
}
return s;
}

private:
int n;
vector<int> c;
};

class Solution {
public:
int createSortedArray(vector<int>& instructions) {
int m = *max_element(instructions.begin(), instructions.end());
BinaryIndexedTree tree(m);
const int mod = 1e9 + 7;
int ans = 0;
for (int i = 0; i < instructions.size(); ++i) {
int x = instructions[i];
int cost = min(tree.query(x - 1), i - tree.query(x));
ans = (ans + cost) % mod;
tree.update(x, 1);
}
return ans;
}
};

• class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)

def update(self, x: int, v: int):
while x <= self.n:
self.c[x] += v
x += x & -x

def query(self, x: int) -> int:
s = 0
while x:
s += self.c[x]
x -= x & -x
return s

class Solution:
def createSortedArray(self, instructions: List[int]) -> int:
m = max(instructions)
tree = BinaryIndexedTree(m)
ans = 0
mod = 10**9 + 7
for i, x in enumerate(instructions):
cost = min(tree.query(x - 1), i - tree.query(x))
ans += cost
tree.update(x, 1)
return ans % mod


• type BinaryIndexedTree struct {
n int
c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
c := make([]int, n+1)
return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) update(x, delta int) {
for x <= this.n {
this.c[x] += delta
x += x & -x
}
}

func (this *BinaryIndexedTree) query(x int) int {
s := 0
for x > 0 {
s += this.c[x]
x -= x & -x
}
return s
}

func createSortedArray(instructions []int) (ans int) {
m := slices.Max(instructions)
tree := newBinaryIndexedTree(m)
const mod = 1e9 + 7
for i, x := range instructions {
cost := min(tree.query(x-1), i-tree.query(x))
ans = (ans + cost) % mod
tree.update(x, 1)
}
return
}

• class BinaryIndexedTree {
private n: number;
private c: number[];

constructor(n: number) {
this.n = n;
this.c = new Array(n + 1).fill(0);
}

public update(x: number, v: number): void {
while (x <= this.n) {
this.c[x] += v;
x += x & -x;
}
}

public query(x: number): number {
let s = 0;
while (x > 0) {
s += this.c[x];
x -= x & -x;
}
return s;
}
}

function createSortedArray(instructions: number[]): number {
const m = Math.max(...instructions);
const tree = new BinaryIndexedTree(m);
let ans = 0;
const mod = 10 ** 9 + 7;
for (let i = 0; i < instructions.length; ++i) {
const x = instructions[i];
const cost = Math.min(tree.query(x - 1), i - tree.query(x));
ans = (ans + cost) % mod;
tree.update(x, 1);
}
return ans;
}